Polynomial Rings and Galois Theory

Polynomial Rings and Galois Theory Giacomo Micheli Balázs Szendrői Introduction Galois theory studies the symmetries of the roots of a polynomial equation. The existence of the two solutions (1) x 1,2 = b ± b 2 4c 2 of the polynomial equation x 2 + bx + c = 0 can be thought of a symmetry of the situation. Assuming that the roots x 1,2 are (real or complex but) irrational, we have the following. Solutions generate a field extension K = Q(x 1, x 2 ) of degree two of the rational field Q. K has a field automorphism (symmetry) defined by φ(x 1 ) = x 2, giving rise to a group G = {Id K, φ} acting on K. φ φ = Id K, so as an abstract group, {e, φ} is of order two, isomorphic to C 2. The rational subfield Q is precisely the subfield of K fixed by both Id K and φ. The main aim of this course is to study groups of automorphisms of field extensions, their fixed subfields, and the relationship between the structure of field extensions and the structure of the associated groups. We will pay special attention to the question whether a general polynomial has roots expressible by a formula consisting of field operations and taking n-th roots, as in formula (1) for the quadratic polynomial. It was known by the 16th century that such a formula exists if deg f 4 (Tartaglia, Cardano, Ferrari). In contrast, it was discovered in the 19th century that there is usually no such formula for deg f 5 (Ruffini, Abel, Galois). We will prove these results as an application of our structure theory of field extensions. 1. Rings, Fields and Polynomial Rings 1.1. Rings and domains. Recall that R is a ring if it has addition, additive inverses, an additive identity 0 R, and it also has multiplication that is distributive over addition. All our rings will also have a multiplicative identity 1 R, and will have both operations commutative and associative. Ring homomorphisms f : R S between rings are required to map the multiplicative identity of R to that of S. For example, Z is a ring while N is not a ring; the map f : Z Z defined by f(n) = 0 is not a ring homomorphism. A ring R is an integral domain, if there are no zero-divisors, i.e. ab = 0 implies that a = 0 or b = 0. A subset I of a ring R is said to be an ideal, denoted I R, if (1) for a R, x I, we have xa I; (2) 0 I and for a, b I, we have a + b, a I. 1

2 Given a ring R and an ideal I, there is a quotient ring R/I consisting of residue classes of elements of R modulo I. Example. Any ideal in the ring of integers is of the form nz = {0, ±n, ±2n,...} Z. The corresponding quotient ring is Z/nZ, which is a domain if and only if it is a field if and only if n is prime. Given any ring R we can define a ring homomorphism φ : Z R such that 1 + 1 +... + 1 (n 0) }{{} n times φ(n) = (1 + 1 +... + 1) (n < 0) }{{} n times Its kernel ker φ = {n Z : φ(n) = 0} is an ideal in Z. So { 0 ker φ = nz. In the first case, Z is a subring of R, and we say R has characteristic 0. In the latter case, Z/nZ becomes a subring of R, and we say that R has characteristic n. In both cases, we call this the prime subring of R. The characteristic of a domain is 0 or a prime number. 1.2. Fields. A ring K is a field if non-zero elements have multiplicative inverses. A field is automatically an integral domain, and therefore has characteristic 0 or prime p. In the first case, we have Q K, in the latter case, F p K, the prime subfield of K. Proposition. Given an integral domain R, there is a field K, the field of fractions of R, with the following properties: there is a homomorphism φ : R K which is injective and for every x K, there are a, b R such that x = φ(a) φ(b). Proof. Construct K as the set of pairs (a, b), a R, b R\{0} under the equivalence relation (a, b) (a, b ) ab = ba. Then K has addition (a, b) + (c, d) = (ad + bc, bd) and multiplication (a, b)(c, d) = (ac, bd). This has all the required structure including multiplicative inverses: (a, b)(b, a) = 1 for a, b 0. Hence K is a field. The homomorphism φ : R K is given by φ(a) = (a, 1) and (a, b) = φ(a) φ(b) if b 0. Lemma. (i) Let K be a field, I K. Then either I = {0} or I = K. (ii) Let K, L be fields. Then if f : K L is a ring homomorphism, then f is injective. Proof. (i) For a 0 I, a 1 K and so 1 I. Then for b K, 1b I and thus K I. So I = K. Part (ii) follows from (i). ker f = {a K : f(a) = 0} K. So either ker f = K, and so f(1) = 0 which cannot happen as f(1) = 1, or ker f = {0} and thus f is injective.

3 1.3. Polynomial Rings. Given a ring R, we can form another ring by considering { d } R[x] = a i x i : a i R, i=0 the set of all polynomials with coefficients in R. This becomes a ring by extending the addition and multiplication in the usual way ( opening the bracket ). For a nonzero polynomial f R[x], its degree deg f is the largest n for which a n 0. A polynomial f R[x] is said to be monic if its leading coefficient is 1. Given a polynomial f K[x], α K is a root of f if f(α) = 0. Lemma. If R is a domain, then R[x] is a domain. Proof. Suppose f, g 0 R[x] with leading coefficients a N x N and b M x M. Then since these are leading coefficients, they are non-zero. But since R is a domain, a N b M 0 and so fg 0. So R[x] is a domain. Now let K be a field. We get a ring K[x], the polynomial ring over K, and its field of fractions { } f(x) K(x) = : f, g K[x], g 0. g(x) Definition. (1) A polynomial f K[x] is said to be irreducible if f = gh only if deg g = 0 or deg h = 0. (2) A polynomial f K[x] divides a polynomial g, written f g, if there exists h K[x] such that g = fh. (3) A polynomial f K[x] is prime, if whenever f gh, (f g) or (f h). An ideal I is said to be prime if ab I implies a I or b I. Hence f is prime in K[x] if and only if (f) is a prime ideal. Recall Proposition. K[x] is a Euclidean ring; that is, given two monic polynomials f, g with deg g 1 then there exist q, r K[x] wth deg r < deg g, satisfying f = gq + r. This implies Theorem. If K is a field, then K[x] is a unique factorization domain. That is, (1) The irreducible polynomials are precisely the prime polynomials. (2) Given f 0 K[x], it can be expressed as a product f = f 1 f 2 f 3...f n of irreducibles in an essentially unique way (up to rearrangement and multiplication by scalars). Corollary. Let K be a field. (1) Given a polynomial f K[x], a K is a root if f(a) = 0 f(x) = (x a)g(x). (2) Given f K[x], there exists distinct roots a i K, multiplicities m i N, and g K[x] without roots, such that f = g(x) i (x a i ) mi. Concerning irreducibility, we have the important and useful

4 Lemma. (Gauss Lemma) Let f Z[x] be monic. Then f is irreducible in Z[x], if and only if f is irreducible in Q[x]. Corollary. Let f be a monic polynomial with integer coefficients and deg f 3. Then if f has no integral root, then f is irreducible in Q[x]. Proof. By Gauss Lemma, f(x) is reducible in Q[x] if and only if f(x) is reducible in Z[x] if and only if f = f 1 f 2 with 1 deg f 1, 1 deg f 2. Since we have made the assumption that deg f = 3, it must be that deg f 1 = 1 (wlog) and so f = (x α)f 2 (x). Then α Z is a root. Proposition. (Eisenstein s Criterion) Let d 1 f(x) = x d + a i x i be a monic polynomial over Z. Suppose that for some prime p, we have that p a i for 0 i < d, but p 2 a 0. Then f(x) is irreducible over Z. We are interested in methods of determining whether monic polynomials are irreducible in Z[x] (and hence in Q[x]). There are various tools that can be used to do this: (1) Eisenstein s criterion. (2) More generally, reduction modulo p, for p a prime. Given a polynomial f Z[x], we have (f mod p) F p [x] = (a i mod p)x i. Clearly, if f = gh then f p gh. Conversely, if for some p, (f mod p) is irreducible, so if f. Checking irreducibility in F p [x] is a finite (though perhaps cumbersome) task, since we have only finitely many choices for each coefficient in a splitting. (3) Substitution: if f(x) = g(x)h(x) then f(x a) = g(x a)h(x a). So f(x a) irreducible implies f(x) is irreducible. There is an example of this on a problem sheet. This might help (say by the fact that f(x a) may be an Eisenstein polynomial). (4) Tricks and ingenuity. 2.1. Basic notions. i=0 2. Group Actions on Rings and Fields Definition. A group G acts on a ring R (on the left) if for every g G, we are given a ring automorphism (bijective ring homomorphism) g : R R mapping a g(a), such that for a R and g, h G, we have h(g(a)) = (hg)(a). (For G to act on R on the right, often written a ag or a a g, we require (ag)h = a(gh)). Lemma. Let the group G act on the ring R. (1) Consider R G = {a R : g(a) = a for all g G}. Then R G is a subring of R. (2) If R = K is a field, then for g G and a K \ {0} we have g(1) = 1 and g(a 1 ) = (g(a)) 1. (3) If R = K is a field, then K G is a subfield, containing the prime subfield of K.

Proof. For (1), just use the subring test. For (2), g(1) = g(1 2 ) = g(1)g(1) so g(1) = 1 so 1 = g(1) = g(aa 1 ) = g(a)g(a 1 ). For (3), taking an element a K G we have that g(a 1 ) = g(a) 1 = a 1 and so a 1 K G. Moreover, 1 is fixed, and the prime subfield consists of elements which are ratios of sums of 1, so the prime subfield must always be fixed. For K a field, denote by Aut(K) the group of all field automorphisms (bijective field homomorphisms) g : K K, with the group operation being composition. More generally, for a field extension L/K, let Aut K (L) = {g Aut(L) for all a K, g(a) = a} denote the group of all field automorphisms of L over K. Using this notation, an action of a group G on a field K is a group homomorphism G Aut(K). In most of our examples, this homomophism will be injective (different elements of G give different automorphisms of K) in which case we will call the action faithful. 2.2. Symmetric Polynomials. Take K a field and let R = K[x 1, x 2,..., x n ] be the polynomial ring in n variables. This is a domain, since we can define this iteratively adding one variable at a time. The ring R has an action by the symmetric group S n by permuting variables: if f(x 1,..., x n ) is a polynomial and σ S n then f σ (x 1,..., x n ) = f(x 1σ,..., x nσ ). (This is a right action; traditionally permutations σ S n act on {1,..., n} on the right). Definition. The ring of symmetric polynomials is the fixed ring K[x 1,..., x n ] Sn. The following elementary symmetric polynomials are clearly elements of the ring K[x 1,..., x n ] Sn : 5 For example. k s k = x ij. i 1<i 2<...<i k j=1 s 1 = i x i, and so on. s 2 = x i x j, i<j s 3 = x i x j x k, i<j<k Theorem. (Theorem on Symmetric Functions) K[x 1,..., x n ] Sn has the following structure. K[x 1,..., x n ] Sn = K[s 1,..., s n ]. In other words, every symmetric polynomial f K[x 1,..., x n ] Sn is uniquely expressible as a polynomial of the elementary symmetric polynomials s i.

6 Proof. We use a trick. Introduce lexicographic ordering on monomials: call the monomial x a1 1... xan n larger than the monomial x b1 1... xbn n, if a 1 > b 1, or a 1 = b 1 and a 2 > b 2 or... Suppose that f K[x 1,..., x n ] Sn. We find the largest monomial term in f in the lexicographic ordering. Let this be i xai i. Since f is symmetric, and i xai i is the largest, we must have a 1 a 2... a n. Notice x ai i is also the largest term in the monomial s ai ai+1 i in the elementary symmetric functions. Now for some c K, g = f c s ai ai+1 i K[x 1,..., x n ] Sn and the largest term in g is smaller than that of f. So we may repeat for g. This shows both that f can be written as a polynomial in the s i and that such a polynomial is unique. Remark. The proof is actually constructive: it gives us an algorithm to write a symmetric polynomial as a polynomial in the elementary symmetric polynomials. In practice, we often use this algorithm for the first few steps and then resort to guesswork and substitutions. Example. The sum of squares i x2 i is not an elementary symmetric polynomial, but it is clearly symmetric. We have x 2 i = s 2 1 2s 2. Proposition. i (1) The discriminant of x 1,..., x n, defined by = i<j(x i x j ) 2, is a symmetric polynomial, an element of K[x 1,..., x n ] Sn. (2) The expression δ = i<j(α i α j ) is not symmetric, but it is invariant under the alternating group A n < S n. Proof. (1) is clear. For (2), note that a transposition (i i + 1) S n multiplies δ by ( 1). So δ is certainly not invariant, but it is invariant for an even product of such transpositions. Such even products generate the alternating group A n. Example. For two variables, we have the well-known expression = (x 1 x 2 ) 2 = s 2 1 4s 2, which is just the b 2 4c in the solution formula of the quadratic polynomial. 3. Field Extensions 3.1. Basic notions. A field extension M/K (or M : K)is just an injection of fields equivalently a field inclusion f : K M, K M

7 which we will often write in diagram form as M K. In particular, if a group G acts on M, then M G M is a subfield or M/M G is a field extension. The degree of a field extension M/K, denoted [M : K], is the vector space dimension dim K M, which may be infinite. M/K is called a finite extension if [M : K] is finite. Recall Proposition. (The Tower Law) If L/M/K are field extensions (finite or infinite) then [L : K] = [L : M][M : K]. Definition/Proposition. We say that a M is algebraic over K, if f(a) = 0 for some non-zero polynomial f K[x]. If a is algebraic over K, its minimal polynomial m a K[x] over K is the unique monic polynomial of least degree satisfying m a (a) = 0. This polynomial has the following properties. (1) m a is irreducible. (2) If f(a) = 0 for f K[x], then m a divides f in K[x]. Definition. A field extension M/K is algebraic, if every element a M is algebraic over K. Otherwise, M/K is transcendental. Note that every finite extension is algebraic, but not every algebraic extension is finite. 3.2. Simple Extensions. Let M/K be a field extension and let a M. The simple extension of K generated by a in M, denoted K(a), is the smallest subfield of M containing K and a. More generally, for a subset S of M, the field K(S) is the smallest subfield of M containing K and S. Example. (1) Let K be a field and K(t) the field of fractions of the polynomial ring K[t]. Then K(t) is a simple transcendental extension of K. (2) Q(i, 2)/Q is a simple algebraic extension of Q, generated by 2 + i. Theorem. (Existence of simple extensions) (1) Given a field K and a monic irreducible polynomial m K[x], there exists a field extension M/K with the following properties: (a) M = K(α) for some element α M. (b) The minimum polynomial of α M is exactly m. (c) [M : K] = deg m. (2) Given L/K and α L algebraic over K, the simple extension K(α) of K in L is isomorphic to the extension constructed in (1) using the minimal polynomial m α of α over K. Proof. (i) m is monic and irreducible, so (m) is a prime ideal in K[x]. quotient ring M = K[x]/(m) So the is a field. Let α M be the image of x in M. Clearly every element of M is a polynomial in α with coefficients in K and so M = K(α). Also, m K[x] maps to 0 M, so m(α) = 0 in M. The minimal polynomial of m α divides m in K[x]. But

8 m is irreducible and so m = m α. Finally, let m = x d + lower order terms K[x]. Then {1, x, x 2,..., x d 1 } is a basis of M over K. (ii) α L is algebraic, and so has a minimal polynomial. Then can define a field homomorphism K[x]/(m α ) L by mapping f(x) f(α), well-defined since m α (α) = 0. So we get M L, and its image is exactly K(α). So we have M = K(α). Theorem. (Extending homomorphisms to simple extensions) Let M/K be a field extension, where α M is algebraic over K with minimal polynomial m α. Let i : K L be a field homomorphism and β L. Then there is a homomorphism j : K(α) L with the following properties: j K = i j(α) = β if and only if i(m α )(β) = 0, where i(m α ) L[x] is the image of m α K[x] under i. Proof. Suppose that such a j exists. Then i(m α )(β) = j(m α )(j(α)) = j(m α (α)) = 0. So β = j(α) is a root of the polynomial m α that had α as a root. So the condition is necessary. Now suppose that i(m α )(β) = 0. Let K = i(k) L. Then i(m α ) K[x] is monic and irreducible, so it is the minimal polynomial of β over K. From the previous Theorem, K(α) = K[x]/(m α ), whereas K(β) = K[x]/i(m α ). Now we can construct the map j as the following composite: K(α) = K[x]/(m α ) i K[x]/i(m α ) = K(β) L. This is a field homomorphism j : K(α) L with j k = i and j(α) = β. Example. (1) Let K = Q and let M = C and let α = i. Then m α = x 2 + 1. Take L = Q(i) and β = i. So i(m α ) = x 2 + 1 and hence i(m α )(β) = 0. So there is a field homomorphism j : Q(i) Q(i) taking i i. (2) Let K = Q, L = Q(i), α = 2, m α = x 2 2 and β = ±i. Can we extend i to K( 2)? No, since i(m α )(±i) = (±i) 2 2 0. Corollary. (Uniqueness of simple algebraic extensions) Suppose we have a field extension M/K with α, β M algebraic over K with the same minimal polynomial m K[x]. Then there is an isomorphism j : K(α) K(β) with j k = Id. Proof. Take i = Id K, L = K(β) in the above theorem. We get K(α) j K(β) and j exists, since m(β) = 0. K Id K K

9 Then j : K(α) K(β) is injective. Also by the Tower Law, hence K(α) = K(β). [K(β) : j(k(α))] [K(α) : K] = [K(β) : K], Corollary. (Homomorphisms from simple extensions) Let K(α)/K be a simple algebraic extension with α having minimal polynomial m α over K. Let i : K L be a homomorphism. Suppose that i(m α ) L[x] has exactly k distinct roots {β i } i k in L. Then there are exactly k distinct homomorphisms j m : K(α) L with the property that j m K = i and these are distinguished by j m (α) = β m for m k. 4.1. Basics. 4. Splitting Fields Definition. Let K be a field, and f K[x] a monic polynomial. Then we say that f splits (completely) over K, if n f(x) = (x a i ) K[x]. i=1 Example. x 2 + 1 does not split over Q, but it does split over Q(i) as x 2 + 1 = (x i)(x + i). Trivially, an irreducible polynomial over K of degree at least 2 does not split over K. Definition. Let K be a field and f K[x]. A field extension M/K is a splitting field of f over K, if (1) f M[x] splits over M; (2) if K L M, then f L[x] does not split over L. Example. The splitting field of x 2 + 1 over Q is Q(i). x 2 + 1 splits over Q(i) and for degree reasons there is no intermediary field. Theorem. (Existence and Uniqueness of Splitting Fields) (1) Given f K[x], f has a splitting field M/K over K. (2) Consider f K[x] and a field isomorphism i : K K taking f i(f) = f K [x]. Given splitting fields M/K for f, and M /K for f, there is an isomorphism j : M M extending i : K K. Proof. We do an induction on the degree of f, noting that all statements are trivial for deg f = 1. For (1), the induction hypothesis is as follows: for any field K, f K[x] with deg f < n, there is a splitting field M/K of f over K. Take f K[x] of degree n, and let f 1 be an irreducible factor of f. Let K 1 := K[x]/(f 1 ). Then this is a finite extension of K. So in K 1 [x] we can write f = (x α 1 ) m1 g(x). Now deg g < deg f and so has by induction a splitting field N/K 1 over K 1. Now in N, g splits and so f splits completely. Take a smallest subfield M of N in which f splits completely; this is a splitting field of f. There were choices in the proof of (1), so the statement of (2) is not obvious. We must prove that the resulting fields are isomorphic.

10 For (2), the induction hypothesis is as follows: for any fields i : K = K, f K[x] with deg f < n, we have that the splitting fields of f and i(f) over K, K are isomorphic. Take any f K[x] of degree n. Take a splitting field M/K for f over K. Let α 1 M be a root of f and let m be the minimal polynomial of α 1 over K. Now f(α 1 ) = 0, so m divides f in K[x]. For m = i(m) K [x], m is irreducible in K [x], and we similarly have that m divides f in K [x]. So m splits in the splitting field L of f over K. Let β 1 be a root of m in M. By uniqueness of simple extensions, the extensions K(α 1 ) and K(β 1 ) are isomorphic, since they correspond to the minimial polynomials m and m = i(m). Now M is the splitting field of f/(x α 1 ) over K(α 1 ) and M is the splitting field of f /(x β 1 ) over the isomorphic field K (β 1 ). So by induction, there is an isomorphism L = L extending K(α 1 ) = K (β 1 ), which itself extends K = K. 4.2. Normal Extensions. Definition. A field extension L/K is normal if, wherever an irreducible polynomial f K[x] has a root in L, then f splits in L. Example. Consider K = Q and the polynomial x 3 2, with the field extension L = Q ( 3 2 ) = Q[x]/(x 3 2). Claim. L/K is not normal. Proof. f has one root α = 3 2 L, but no other roots in L, since all other roots of f over Q are non-real. The splitting field M of f is not contained in the real numbers: ( ) 3 M = Q 2, ω, with a third root of unity. ω = e 2πi 3 Example. The field extensions Q(i)/Q, Q( 2)/Q are normal, since if a quadratic splits into factors then it splits completely into linear factors. Theorem. (Characterisation of Normal Extensions) A finite extension L/K is normal if and only if it is the splitting field of some f K[x]. Proof. Assume first that L/K is normal and finite. If L = K then there is nothing to prove. Otherwise, take α 1 L \ K, α 2 L\K(α 1 ), and so on; since the degree increases in each step, this process must terminate, with a generating set L = K(α 1,..., α n ), with each α i algebraic over K. Let m i K[x] the minimal polynomial of α i over K. Set f = m i K[x]. Thus each m i has one root α i in L, and as L/K is normal, it splits in L. Hence f splits in L. Furthermore, any splitting field of f over K must contain all roots of m i and so no subfield of L splits f. Conversely, suppose that L is the splitting field of f K[x]. Take g K[x] irreducible, with a root α 1 L. Consider a splitting field M L of fg over K.

11 Let α 2 M be another root of g in M. We want that α 2 L. We argue using the following diagram. M L(α 1 ) L(α 2 ) L K(α 1 ) K(α 2 ) K We have K(α 1 ) = K(α 2 ) and [K(α 1 ) : K] = [K(α 2 ) : K] = d = deg g, since both are simple extensions using the irreducible g K[x]. Next, L(α i ) is the splitting field of f over K(α i ). Thus, by Uniqueness of Splitting Fields, we get an isomorphism L(α 1 ) = L(α 2 ) extending K(α 1 ) = K(α 2 ). In particular, [L(α 1 ) : K(α 1 )] = [L(α 2 ) : K(α 2 )] = e for some integer e. Recalling α 1 L, we have [L(α 1 ) : L] = 1. Hence by the tower law, Hence [L : K] = [L(α 1) : K(α 1 )] [K(α 1 ) : K] [L(α 1 ) : L] = d e. [L(α 2 : L)] = [L(α 2) : K(α 2 )] [K(α 2 ) : K] [L : K] = d e d e = 1. Thus α 2 L. This completes the proof. 4.3. Separable extensions. Definition. An irreducible polynomial f K[x] is separable, if it has deg f distinct roots in a splitting field. An arbitrary polynomial f K[x] is separable if all its irreducible factors are. Definition. Given a field extension L/K, α L is separable over K, if its minimal polynomial m α K[x] over K is a separable polynomial. A field extension L/K is separable, if every α L is separable over K. Theorem. (Separability in characteristic 0) Assume that char K = 0. Then any finite extension L/K is separable. Proof. Let m K[x] be the minimal polynomial of α L over K. Suppose that m has a double root β in some splitting field M K. Then m(β) = Dm(β) = 0, where D is the formal derivative (see Problem Sheet 1). On the other hand, Dm is a polynomial with co-efficient of x d 1 nonzero since char K = 0, hence Dm is nonzero. m K[x] is irreducible, and Dm has smaller degree, so (m, Dm) = 1 and thus there are a, b K[x] such that am + b(dm) = 1. But substituting β we get a contradiction.

12 Example. Take K = F p (t). Then this is an infinite field of characteristic p. Consider the polynomial f(x) = x p t K[x]. Let L/K be a splitting field of f and let α L be one root of f. Then in L[x], we have that (x p t) = (x α) p (since β β p is a field homomorphism in characteristic p). So f is a non-separable polynomial in K. The proof of the Theorem breaks down, since Df = px p 1 = 0 since char K = p. Proposition. (Separability of intermediate extensions) If M/K is separable, then for any intermediate field L, the extensions M/L and L/K are separable. Proof. L/K is clearly separable, since the minimal polynomial of α L over K is the minimal polynomial of α M over K. On the other hand, the minimal polynomial m L[x] of β M over L divides the minimal polynomial m K[x] of the same element β M over K, by properties of the minimal polynomial. If m has no roots in a splitting field, neither has m. Theorem. (Extending homomorphisms to separable extensions) Let M/K be a field extension of degree d. Let i : K L be a field homomorphism to some other field L. Then there exist exactly d homomorphisms j k : M L extending i Otherwise, there are fewer than d extensions of i. M/K is separable, and the minimal polynomial. of every α M splits in L[x] Proof. We work by induction on d, the case d = 1 being obvious. Let d > 1, and assume that the stated equivalence is true for all extensions M/K of smaller degree. Suppose first that (1) either M/K is not separable, (2) or there is α M/K whose minimal polynomial over K does not split in L. In both cases, for some α M/K its minimal polynomial has fewer than deg(m) distinct roots in L. So there are fewer than deg(m) extensions of i : K L to K(α) L. Also, by induction, there are at most [M : K(α)] extensions of each such j to j : M L. Hence, there are fewer than deg(m) [M : K(α)] extensions of i : K L. However, this number is deg(m) [M : K(α)] = [K(α) : K][M : K(α)] = [M : K] = d. Now assume that M/K is separable and that all m α split in L[x], α M. Take α M\K, and let m be its minimal polynomial over K. Then m L[x] splits into distinct linear factors m(x) = (x β k ) for β k L different. deg m k=1 Now by the Theorem on extending homomorphisms to simple extensions, we have deg(m) distinct extensions of i to K(α). By induction, each of these in turn leads [M : K(α)] extensions of i, giving a total of [M : K(α)] deg(m) = d extensions.

13 Corollary. The splitting field L over K of a separable polynomial f K[x] is a separable extension of K. Remark. The statement is not obvious: for L/K to be separable, we need all minimal polynomials to be separable, not just (components of) f. Proof. We will give a sketch proof here; see the lectures for more details. Let K be a subfield of a field M and let i : K L be a homomorphism of fields. By the proof of the last theorem, using induction on d = [M : K], we have (i) if M : K is not separable, or if there is some α M such that i(m α ) does not split in L, then there are strictly fewer than d field homomorphisms j : M L such that j K = i; and (ii) if M = K(α 1,..., α n ) where for 1 k n the minimal polynomial m αk of α k over K is separable and its image under i splits in L, then there are exactly d field homomorphisms j : M L such that j K = i. Now let M = L be the splitting field of a separable polynomial f over K. Then M = K(α 1,..., α n ) where α 1,..., α n are the roots of f in L, and for 1 k n the minimal polynomial m αk of α k over K divides f, so m αk is separable over K and its image under the inclusion i : K L splits in L. Thus (ii) above is satisfied and therefore by (i) the extension is separable. 5. Galois Extensions and their Galois Groups Example. Consider the field extension Q( 2, i)/q. Our aim is to find and understand field automorphisms of the field Q( 2, i) that fix the subfield Q, in the hope that this will help us to understand the structure of the extension. The field Q( 2, i) is the splitting field of the irreducible polynomial x 2 + 1 over Q( 2); the elements ±i both have minimal polynomial x 2 + 1 over this field. Therefore, by the Theorem on extending field homomorphisms to simple extensions, there is a field automorphism Q( σ 2, i) Q( 2, i) Q( 2) = Q( 2), with σ : i i. We can also regard Q( 2, i) is the splitting field of x 2 2 over Q(i). So using the same argument, there is a field automorphism τ defined by Q( 2, i) τ Q( 2, i) Q(i) = Q(i), with τ : 2 2. Consider the group G =< σ, τ >, which acts on the field Q( 2, i) fixing Q. The actions of elements of G on a basis of Q( 2, i) over Q can be tabulated as follows. Notice that σ 2 = Id τ 2 = Id (στ) 2 = Id

14 1 2 i 2i Id 1 2 i 2i σ 1 2 i 2i τ 1 2 i - 2i στ 1 2 i 2i and therefore as an abstract group, G = C 2 C 2. Finally, let us consider the subgroups of the group G. We get the following structure, where lines denote inclusions of subgroups. {id} σ τ τσ G Correspondingly, there is a diagram of subfields of Q( 2, i): Q( 2) Q( 2, i) Q(i) Q( 2i) Q The diagrams contain all subgroups, respectively subfields, and each subfield is the fixed field of the subgroup in the same position in the table. Example. Consider Q( 3 2)/Q. Are there any field automorphisms of Q( 3 2) fixing Q? Note that a field automorphism takes a root of an irreducible polynomial f Q[x] to another root of the same irreducible polynomial f since coefficients of the minimal polynomial are fixed by σ. Applying this to the minimal polynomial f(x) = x 3 2, we get that for any σ Aut(Q( 3 2)), σ( 3 2) is a root of x 3 2. However, there is only one root of x 3 2 in this field. So σ( 3 2) = 3 2 and so σ = Id. The problem here is that the field extension Q( 3 2)/Q is not normal. Example. Let L be the splitting field of x p t over F p (t), which is given by F p ( p t) : F p (t). Are there any automorphisms of L fixing K = F p (t)? This polynomial has only one root with multiplicity, so again any automorphism is the identity. The problem here is that the field extension L/F p (t) is not separable. 5.1. Definitions. Recall that for a field extension L/K, we defined Aut K (L) = {α Aut(L) : α K = Id K }. From now on, we call this the Galois group of L over K, and denote it by Γ(L/K), or sometimes G(L/K). Remark. (1) Γ(L/K) is a subgroup of Aut(L). (2) Γ(L/K) acts on L fixing K. (3) By definition, the fixed field L Γ(L/K) contains K, but there is no a priori reason why these two fields should be equal.

15 Definition. The field extension L/K is a Galois extension, if L Γ(L:K) = K. Example. Let K = Q( 2, i) and G = Γ(Q( 2, i)/q) = C 2 C 2. As we saw above, the fixed field K G = Q and so K/Q is a Galois extension. Colloquially we say that the extension is Galois. Γ(Q( 3 2)/Q) = {Id} and Q( 3 2) {Id} Q, so Q( 3 2)/Q is not Galois. Similarly, the splitting field L of x p t over F p (t) is not Galois. The following Theorem summarizes our discussion through examples above. We will prove it in stages in the following sections. Theorem. (1) A finite extension L/K is Galois if and only if L/K is normal and separable. (2) Suppose that L/K is Galois. Then there is an inclusion reversing bijection between {Intermediate fields L M K} and {Subgroups of Γ(L : K)} given by M {σ Aut(L) : σ M = Id M } L H H. 5.2. Field degrees and Group Orders. This section is the techinical core of our proof of the Fundamental Theorem. Lemma. (Linear Independence of Field Homomorphisms) Let L, K be fields and let σ k : K L be distinct field homomorphisms k = 1,..., n. Then {σ 1,..., σ k } are linearly independent over L. Proof. Suppose the converse. Then m i=1 λ iσ i 0 be the shortest relation (m n) with λ i L \ {0}. So for x K, m i=1 λ iσ i (x) = 0. Pick α K such that σ 1 (α) σ 2 (α). Then for x K, m λ i σ i (αx) = 0 i=1 m λ i σ i (α)σ i (x) = 0 i=1 m λ i σ 1 (α)σ i (x) = 0 i=1 m λ i (σ i (α) σ 1 (α))σ i (α) = 0. i=2 So with µ i = λ i (σ i (α) σ 1 (α)), is a shorter relation. m µ i σ i 0 i=2 Recall that a group G acts on a field L faithfully if different group elements give different field automorphisms; in other words, the map G Aut(L) is injective.

16 Theorem. (Degree of fixed subfield) Suppose a finite group G acts faithfully on a field L. Then [L : L G ] = G, the number of elements of G. Proof. Let G = {g 1 = Id L, g 2,..., g n }, with g i : L L field automorphisms. Let m = [L, L G ] (perhaps m = ). Suppose first that m < n. Let {α 1, α 2,..., α m } be a basis of L over L G. For n unknowns β k L, consider the system of m equations n g k (α i )β k = 0 for i = 1,..., m. k=1 Since n > m, there are more unknowns than equations. Hence by the Rank-Nullity Theorem, there is a nonzero solution {β k }. Hence, for α L, n β i g i (α) = 0, and so i=1 n β i g i 0, i=1 as a map L L. This contradicts Linear Independence of Field Homomorphisms. Now assume n < m so that there is a linearly independent subset {α 1,..., α n+1 } of L over L G with n + 1 elements. This time consider the set of n equations n+1 g j (α i )β i = 0 for j = 1,..., n, i=1 for n + 1 unknowns β i L. Once again, there are more unknowns than there are equations, so there is a nonzero solution. Reorder this solution {β 1,..., β n+1 } such that β i,..., β r 0 and β r+1,..., β n+1 = 0, and choose a solution with minimal r 1. So r (2) g j (α i )β i = 0. i=1 Operate on this by an element g G: r gg j (α i )g(β i ) = 0 i=1 which implies (through renaming group elements) r (3) g j (α i )g(β i ) = 0. i=1 Now take (2) g(β 1 ) (3) β 1 to get r g j (α i )(β i g(β 1 ) β 1 g(β i )) = 0. i=2 The minimality of r implies that all coefficients here are zero. So for all g G, that is, β i g(β 1 ) β 1 g(β i ) = 0, β i /β 1 = g(β i /β 1 )

17 and therefore But from (3), we deduce that β i /β 1 L G. r i=1 α i β i β 1 = 0 which gives a contradiction, since {α 1, α 2,..., α n+1 } was chosen to be linearly independent over L G. Corollary. If a finite extension L/K is Galois, then the number of elements in the Galois group Γ(L/K) is the same as the degree [L : K]. Proof. This follows from the previous Theorem, together with the Galois property. 5.3. Characterisation of Galois Extensions. Theorem. (Characterisation of Galois Extensions) A finite extension L/K is Galois if and only if L/K is separable and normal. Proof. Suppose that L/K is separable and normal of degree n. Then by the Theorem on extending homormophisms to separable extensions, there are exactly n = [L : K] homomorphisms L L extending the inclusion K L. Thus Γ(L/K) = n. Hence [L : L Γ(L:K) ] = n. But L Γ(L/K) K, [L : L Γ(L/K) ] = n = [L : K] which implies K = L Γ(L/K). Hence L/K is Galois. Conversely, suppose that L/K is Galois of degree m. Then there exist exactly m homomorphisms L L extending the inclusion K L, the elements of Γ(L : K). So by the converse direction of the Theorem on extending homormophisms to separable extensions, L/K is separable and the minimal polynomial of every α L over K splits in L. So L/K is normal. 5.4. The main theorem. Theorem. (Fundamental Theorem of Galois Theory) Suppose that L/K is finite and Galois, with Galois group G. (1) There is an inclusion reversing correspondence between intermediate field extensions L/M/K and subgroups of G, given as follows: { } intermediate fields {subgroups H < G} L M K The degrees are given by M G M = {α Aut(L) : α M = Id M } L H H. [L : M] = G M, [M : K] = G G M. (2) The intermediate field extension M/K is Galois if and only if G M G is a normal subgroup. In this case, its Galois group is given by Γ(M/K) = G/G M.

18 Proof. For (1), let M be an intermediate field. Then L/M is normal (it is a splitting field over K and so a splitting field over M). Also, L/K separable implies L/M is separable by the Proposition on Separability of intermediate extensions. It follows that L/M is Galois with Galois group Γ(L/M) = G M, acting on L as a subgroup of G, with fixed field L G M = M. Conversely, consider a subgroup H < G. It is clear that H G L H : H is contained in the group that fixes all elements fixed by H. Now by what we proved in the previous step, we have that L G L H = L H. On the other hand, by the theorem on fixed fields and group orders, H = [L : L H ] = [L : L G L H ] = G L H. So H and G L H are both subgroups of G, H G L H and they have the same size. So H = G L H. Hence the maps M G M and H L H are mutual inverses, so indeed we have a one-to-one correspondence. Clearly, as M gets larger, G M gets smaller and vice versa. So the correspondence is inclusion reversing. To do (2), we need a lemma. Lemma. Let L/K be a field extension, L M K, τ Γ(L/K) = G. Then G τ(m) = τg M τ 1, as subgroups of G. Slogan: Stabilizer of a translate is the conjugate of the stabilizer. Proof. If α M, g G M with gα = α then implies The converse works likewise. (τgτ 1 )(τα) = τgα = τα τgτ 1 G τ(m). Recall we have L M K. Suppose that M/K is Galois, in particular normal. Given g G and α M let m be the minimal polynomial of α over K. Then m(g(α)) = g(m(α)) = 0, so g(α) is another root of the irreducible polynomial m K[x]. m has one root α M, and so by normality it splits in M. So g(α) M. Therefore, g(m) = M for all g G. By the Lemma therefore, for all g G, gg M g 1 = G M. So G m G is a normal subgroup. Conversely, suppose G M is a normal subgroup of G. We want to show that the corresponding extension M/K is Galois. Let α M, β L a different root of the minimal polynomial m of α over K. Then K(α) is a subfield of L and K(β) is a subfield of L, so there are isomorphisms of fields K(α) = K(β) with α β. By the theorem on uniqueness of splitting fields, this extends to an isomorphism L = L K(α) = K(β),

19 in other words an element g Γ(L/K), such that g(α) = β. But G g(m) = G M by the lemma and the normal subgroup property, so g(m) = M, by the one-to-one correspondence of (1). So β M. Hence M/K is normal. Separability follows from that of L/K. M/K is thus Galois. Finally, if M/K is Galois, define a map σ : G Γ(M/K) by σ(g) = g M. This is surjective by the extension theorem, with ker σ = G M. The isomorphism theorem for groups now gives as claimed. Γ(M/K) = G/G M 6. Galois groups of polynomials 6.1. Basics. Take f K[x], and for simplicity assume char K = 0 in this section. Let L/K be a splitting field of f over K, and let G f := Γ(L/K). Note that, since we assumed char K = 0, L/K is automatically separable and hence Galois. Recall that if n 1 f = x n + b i x i K[x] = i=0 n (x α i ) L[x] i=1 then the coefficients b i K are symmetric polynomials in the roots α i L. In particular, the discriminant f = (α i α j ) 2 i<j of f is a polynomial in the b i s. Finally recall that δ f = i<j(α i α j ) is not symmetric, but it is invariant under A n < S n. Proposition. (1) If f K[x] has n distinct roots then G f is naturally a subgroup of the symmetric group S n. (2) If f is irreducible, then G f is a transitive subgroup of S n. (3) If f K[x] has degree n and has n distinct roots then G f is contained in the alternating group A n if and only if f K 2 = {a 2 : a K}. Proof. (1) If α L is a root of f K[x], then for all g G f, g(f(α)) = f(g(α)) = 0 and so g(α) L is another root of f. If α 1,..., α n are all the roots of f, then L = K(α 1,..., α n ). So we obtain a homomorphism G S n with g (α i α j ). This homomorphism is injective, since g fixes K and g : α i α j determines the action of g on L.

20 (2) Using the argument often used before, if f irreducible, then for α i, α j different roots, there is an isomorphism K(α i ) = K(α j ) extending to an automorphism g : L L such that g(α i ) = α j. So G f is transitive. (3) We have f K 2 if and only if δ f K if and only if δ f L G f = K if and only if G f A n, by the last observation before the statement of the proposition. Lemma. Let L be a finite field, then L = L \ {0} is cyclic. Proof. Recall that a finite group is cyclic if and only if there is exactly one subgroup for each divisor of the group order. Suppose L is not cyclic and then fix two subgroups G 1, G 2 having the same order d. Consider the polynomial f = x d 1. The set of roots of this polynomial contains S = G 1 G 2. Since S has size greater or equal than d + 1, we get the contradiction, as L is a field f has degree d. Theorem. (Theorem of the Primitive Element) If L/K is finite and separable, then there is some θ L such that L = K(θ). Example. Q( 2, i) = Q( 2 + i). Proof. If K is finite, than L is finite and we can set θ to be the generator of L = L \ {0}. Therefore, we can assume that K is infinite. Since L is finite dimensional over K, it can be written as K(α 1,..., α n ) for some n N and α 1,..., α n M. If we can find θ 1 such that K(α 1, α 2 ) = K(θ 1 ), the claim will follow by induction. Let m 1 and m 2 be the minimal polynomials of α 1 and α 2 respectively. Consider the splitting field of m 1 m 2, which is a Galois extension of K. Using the fundamental theorem of Galois Theory we observe that the number extensions of K contained is M is finite (why?). Therefore, the set of extensions of the form {K(α 1 +bα 2 )} b K is finite. Which directly implies K(α 1 + cα 2 ) = K(α 1 + dα 2 ), which forces α 1 + cα 2 K(α 1 + dα 2 ), from which it follows α 2 K(α 1 + dα 2 ) and so α 1 K(α 1 + dα 2 ). Therefore, set θ 1 = α 1 + dα 2. 6.2. Polynomials of Low Degree. Quadric Polynomials Let f(x) = x 2 + ax + b, which we reduce to g(y) = y 2 + c using the change of variables x y + a 2. We have f = g = 4c. Then f is irreducible if and only if c / K 2, if and only if [L : K] = 2, and in this case, the Galois group has one nontrivial element σ : c c α 1 α 2 leading to Cubic Polynomials G f = σ / σ 2 = id = S 2. Proposition. Let f K[x] be an irreducible cubic with discriminant, splitting field L/K and Galois group Γ(L/K) = G f. Then (1) if K 2, then G f = A3 and L/K is a cubic extension, generated by any of the roots of f.

21 (2) If K \ K 2, then G f = S3, and the structure of intermediate fields is as follows: with α i the roots of f. L K( ) K(α 1 ) K(α 2 ) K(α 3 ) K Proof. Since f is irreducible, the simple extension it generates is of degree 3. By general theory, G f < S 3 and has order divisible by 3, so it can only be S 3 or A 3, depending on the discriminant. The rest of the statement follows immediately. Example. (1) Consider f(x) = x 3 2 over Q. f is irreducible by Eisenstein, and we have L = Q( 3 2, ω) 3/ \2 Q(ω) 6 Q( 3 2) 2\ /3 Q So G f = [L : Q] = 6 and G f = S3. In this case, we can in fact construct the elements of G f explicitely. We have roots 3 2, 3 2ω, 3 2ω 2 and there is a Galois automorphism σ, fixing Q(ω), and mapping 3 3 2 2ω 3 3 2ω 2ω 2. 3 2ω 2 3 2. There is also an automorphism τ fixing Q ( 3 2 ), and mapping ω ω 2. The elements σ and τ generate a group isomorphic to S 3, with σ acting as a three-cycle on the roots and τ as a two-cycle. (2) f(x) = x 3 x 1 3 Q[x]. Then by Sheet 1, ( = 4( 1) 3 27 3) 1 2 = 4 3 = 1 Q 2. f is irreducible over Q, since f has no root over Q. Thus G f = A3. Quartic Polyomials We have done one example in the introduction to Chapter 5, Q( 2, i) = Q( 2 + i) over Q. Here we obtained Γ(Q( 2, i) : Q) = C 2 C 2 S 4. For another example, let f(x) = x 4 2. Set α = 4 2, then the splitting field is L = Q(α, i). We have [Q(α) : Q] = 4, and [Q(α, i) : Q(α)] = 2 since the two fields cannot be the same (one real, the other not) but the degree is clearly at most two.

22 On the other hand, much as before, there is a Galois automorphism σ, fixing Q(i), and mapping α to iα, as well as an automorphism τ fixing Q(α), and mapping i to i. These elements σ and τ generate a group of order eight, isomorphic to the dihedral group D 8. For further details, read Chapter 12 of Stewart s book. 7. Some special classes of extensions 7.1. Cyclotomic Extensions. Let L be the splitting field of f(x) = x n 1 over a field K, assuming that char K n. Proposition. L/K is Galois. Proof. L is a splitting field, so L/K is normal. The polynomial x n 1 is separable over K, since Df = nx n 1 0 since char L n. So L/K is Galois. Definition. Let Note that µ n (L) is a group. µ n (L) = {α L : α n = 1}. Fact. The group µ n (L) cyclic of order n. You know this fact over Q from the Argand plane description of C; we will take this for granted over arbitrary fields of characteristic prime to n. Definition. An element ω µ n (L) is primitive or a primitive root of 1, if it generates the cyclic group µ n (L). Note that if ω is primitive, then L = K(ω). Note also that if ω is a primitive root, then the other primitive roots are ω i for i U(Z n ), the units in Z n. Theorem. Fix a primitive root ω L of 1. (1) Define Φ n (x) = i U(Z n) (x ω i ). (Note that the product is over all primitive roots of 1.) Then Φ n (x) K[x]. (2) There is an injective homomorphism θ : Γ(L : K) U(Z n ). (3) θ is an isomorphism if and only if Φ n is irreducible. Proof. Let G = Γ(L : K) be the Galois group, then σ G must map a generator of µ n (L) to another generator, and a root of x m 1 to another root. So σ(ω) = ω θ(σ), where θ(σ) U(Z n ). Then this defines θ : G U(Z n ) which is easily seen to be a group homomorphism. It is injective, since the effect of σ on L is determined by the effect of σ on ω, since as we saw, L = K(ω). Now G permutes {ω j : j U(Z n )} so the coefficients of Φ n (x) are fixed by G, and so Φ n (x) L G [x] = K[x]. Φ n is irreducible if and only if ω can be mapped to all other primitive roots, if and only if θ is an isomorphism. Corollary. G is abelian. Proof. Remark. In fact, the cyclotomic polynomial Φ n is irreducible in Q[x] for all n. This is very easy to see for n = p prime. We will not need the general case. Φ n is often reducible over finite fields; see Worksheets for examples.

23 7.2. Kummer Extensions. Continuing our study in the last section, let L be a field in which x n 1 splits. Let us study the polynomial x n θ L[x], for θ L. In other words, we are studying extensions under the assumption that charl n. M = L( n θ), Theorem. Suppose that L is a field with charl n and assume that x n 1 splits in L. (1) Let M be the splitting field of x n θ over L. Then M/L is Galois, with Galois group cyclic of order dividing n. (2) Suppose that M/L is a Galois extension of L with Galois group cyclic of order n. Then there exists θ L and a root β M of x n θ L[x], an irreducible polynomial such that M = L(β). Proof. (i) Let L(β) be a primitive extension of L with a root of x n θ. Let ω L be a primitive n th root of unity. Then in L(β), x n θ = (x β)(x ωβ)(x ω 2 β)... (x ω n 1 β) L(β)[x] so M = L(β) is the splitting field. β M is separable over L and so M/L is separable (see Worksheet 3). So M/L is Galois. Now let G = Γ(M : L), then for σ G, σ(β) = ω j(σ) β, since it has to be another root of x n θ. We get an injective map It is a group homomorphism, since j : G Z n. τ(σ(β)) = τ(ω j(σ) β) = ω j(σ) τ(β) = ω j(σ) ω j(σ) β = ω j(σ)+j(σ) β. Hence G < Z n, so is cyclic of order dividing n. (ii) Let G =< σ > be the cyclic Galois group of M/L. Then id, σ, σ 2,..., σ n 1 : M M are distinct field homomorphisms, so they are linearly independent, hence α M such that (4) β = α + ωσ(α) + ω 2 σ 2 (α) + + ω n 1 σ n 1 (α) 0. So σ(β) = σ(α) + ωσ 2 (α) + ω 2 σ 3 (α) + + ω n 1 α = ω 1 β. Hence β / L (not fixed by σ), σ(β n ) = β n, so β n = θ M G = L. So L(β)/L splits x n θ over L, and < σ > consists of distinct field automorphisms of L(β) fixing L. That is, and so n = [M : L] [L(β) : L] n M = L(β).

24 7.3. Solving cubics by radicals. Let f K[x] be an irreducible cubic, and assume chark 2, 3. Recall the discriminant f of f, and that if L is a splitting field of f over K, then { A3, if K Γ(L : K) = 2 S 3, if / K 2. Let ω be a primitive cube root of unity over K. Then L(ω) is Galois over K(ω, ), with Γ(L(ω) : K(ω, )) having order 3. Since the only group of order 3 is C 3, the Galois group is cyclic. So by the Theorem above, L(ω) is a Kummer extension, the splitting field of some polynomial x 3 θ over K(ω, ). So all the roots of f are contained in the extension K(ω,, 3 θ) of K. Explicitely, take f(x) = x 3 + px + q K[x]. Suppose that the roots are α 1, α 2, α 3. Imitating (4) above, set β = α 1 + ωα 2 + ω 2 α 3 and γ = α 1 + ω 2 α 2 + ωα 3. Explicit calculations show that βγ = 3p β 3 + γ 3 = 27q. So β 3, γ 3 are roots of X 3 +27qX 27p 3 = 0 and we can solve for the roots explicitly from this, recalling that α 1 + α 2 + α 3 = 0. 8.1. Solvable Groups. 8. Solvable Groups and Solubility by Radicals Definition. A finite group G is soluble or solvable, if it has a series of subgroups {e} = G 0 < G 1 < < G n = G, such that G i 1 G i is normal, and G i /G i 1 is abelian. Examples. (1) Any abelian group is solvable; take {e} G. (2) S 3 is not abelian, but it is solvable: take the chain {e} < A 3 < S 3. The quotients are A 3 /{e} = C 3 and S 3 /A 3 = C2. (3) S 4 is not abelian, but still solvable. This time take the chain {e} < V 4 < A 4 < S 4, where V 4 is the Klein Four group generated by all permutations of cycle type 2 2. The quotients are V 3 /{e} = C 2 C 2, A 4 /V 3 = C3 and of course S 4 /A 4 = C2. (4) The group A 5 is not soluble. See below. Proposition. (1) Suppose that G is solvable, H G. Then G is solvable. (2) Suppose G is solvable, N G. Then G/N is solvable. (3) Suppose N G and N, G/N are both solvable. Then G is solvable. Remark. Motto: The set of solvable finite groups is closed under taking subgroups, quotient groups and extensions. This would not be true for abelian groups!

25 Proof. For (i), let {G i } i n be a series for G, and H G. Then let H i := G i H. Clearly H i 1 H i, and the inclusion H i = H G i G i defines, after taking quotients, an injective map H i /H i 1 = H G i /H G i 1 G i /G i 1. The latter is abelian by assumption. So H i /H i 1 is a subgroup of an abelian group and thus abelian. For (ii), let {G i } i n be a series for G, and let N i := G i N/N < G/N. It s clear that N i 1 N i and for the quotient, G i /G i 1 N i /N i 1 = (G i N)/N (G i 1 N)/N g g.e + (G i 1 N)/N is a surjective map. By assumption, G i /G i 1 is abelian, hence N i /N i 1 is a quotient of an abelian group and thus abelian. For (iii), suppose we have a series {N i } i m for N. A result from elementary group theory says that any subgroup of G/N can he written as G i /N for some G i G. So we take the series {G i /N} i n. We then stick the two chains together: N 0 < N 1 < < N m = G 0 < G 1 < < G n. By assumption, each group here is normal in the next one, and successive quotients are N i /N i 1, which is abelian, or G i /G i 1 = G i /N G i 1 /N by the 3rd Isomorphism Theorem, which is also abelian, by assumption. Proposition. The groups A n and S n, for n 5, are not soluble. Proof. As proved in Part A, the group A 5 is simple: it has no normal subgroups at all. Since A 5 is certainly not abelian, it cannot be soluble. The groups A n and S n, for n 5, have A 5 as a subgroup, so they cannot be soluble either. 8.2. Solubility by Radicals. In order to avoid assumptions on the characteristic, assume in this section that all fields have characteristic 0. Definition. A field extension L/K is radical, if there is a sequence of elements α 1,..., α n L and positive integers m 1,..., m n, so that L = K(α 1,..., α n ), and for all i, α mi i K(α 1,..., α i 1 ). Given a polynomial f K[x], we say that f can be solved by radicals, if there is a radical extension M/K in which f splits. Note that M can be larger than the splitting field. Theorem. Suppose L/K is a finite Galois extension. Then there exists a finite extension M of L such that M/K is radical if and only if Γ(L : K) is solvable.